Question

complete the following statement. use the integers that are closest to the number in the middle.
\boxed{} < \sqrt3{120} < \boxed{}

Explanation:

Step1: Find nearest lower cube

$4^3 = 4\times4\times4 = 64$

Step2: Find nearest higher cube

$5^3 = 5\times5\times5 = 125$

Step3: Compare cube root to integers

Since $64 < 120 < 125$, take cube roots: $\sqrt[3]{64} < \sqrt[3]{120} < \sqrt[3]{125}$, so $4 < \sqrt[3]{120} < 5$.

Answer:

$4 < \sqrt[3]{120} < 5$